(a) equal
Explanation: when the two spheres are connected through the wire, the charge on the spheres redistribitute such that the two spheres reach the same potential V. This occurs because if the potential on the two spheres was different, then there would be a net electric field that keeps the charges moving from one sphere to another, until an equilibrium condition is reached (and this equilibrium condition is exactly when the potentials on the two spheres are equal).
(b) [tex]Q_1 = \frac{Q}{9}[/tex]
As stated in the previous exercise, the potential on the two spheres is equal:
[tex]V_1 = V_2[/tex] (1)
The capacity of a charged sphere is given by:
[tex]C=4 \pi \epsilon_0 R[/tex] (2)
where R is the radius of the sphere. Let's also remind the relationship between charge Q, capacity C and potential V:
[tex]V=\frac{Q}{C}[/tex]
So we can rewrite (1) as
[tex]\frac{Q_1}{C_1}=\frac{Q_2}{C_2}[/tex]
and by re-writing the capacity using (2):
[tex]\frac{Q_1}{4\pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2}[/tex] (3)
The problem tells us that sphere 2 has a radius 8 times the radius of sphere 1:
[tex]R_2 = 8 R_1[/tex]
Also, the total charge is [tex]Q=Q_1 + Q_2[/tex], which can be rewritten as
[tex]Q_2 = Q-Q_1[/tex]
By using these last 2 equations into (3), we have:
[tex]\frac{Q_1}{4\pi \epsilon_0 R_1}=\frac{Q-Q_1}{4 \pi \epsilon_0 (8 R_1)}\\8Q_1 = Q-Q_1\\9Q_1 = Q\\Q_1 = \frac{Q}{9}[/tex]
(c) [tex]Q_2 = \frac{8}{9}Q[/tex]
Since the first sphere has a charge of [tex]Q/9[/tex], and the total charge must be Q, the second sphere will have a charge of
[tex]Q_2 = Q-Q_1 = Q-\frac{Q}{9}=\frac{8}{9}Q[/tex]
(d) 8
The surface of a sphere is given by
[tex]A=4 \pi R^2[/tex]
So, the surface charge density of sphere 1 is
[tex]\sigma_1 = \frac{Q_1}{A_1}=\frac{Q/9}{4 \pi R_1^2}=\frac{Q}{36 \pi r_1^2}[/tex]
the surface charge density of sphere 2 is
[tex]\sigma_2 = \frac{Q_2}{A_2}=\frac{\frac{8}{9}Q}{4 \pi (8R_1)^2}=\frac{Q}{288 \pi R_1^2}[/tex]
So, the ratio of the two surface charges is
[tex]\frac{\sigma_1}{\sigma_2}=\frac{\frac{Q}{36 \pi R_1^2}}{\frac{Q}{288 \pi R_1^2}}=8[/tex]
